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LpL_p space

LpL_p space
Let (X,A,μ)(X,\mathcal{A},\mu) be a measure space and 1≤p<∞1\leq p<\infty. The space Lp(X,μ)L_p(X,\mu) consists of measurable functions f:X→Rf:X\to\mathbb{R} satisfying
∥f∥p:=(∫X∣f(x)∣p dμ(x))1/p<∞.\begin{equation*}\|f\|_p := \left(\int_X |f(x)|^p\,\mathrm{d}\mu(x)\right)^{1/p}<\infty.\end{equation*}
Essential boundedness and L∞L^{\infty} space
A measurable function f:X→Rf:X\to\mathbb{R} is essentially bounded if there is a constant 0≤K<∞0\leq K<\infty such that ∣f(x)∣≤K|f(x)|\leq K for μ\mu-almost every x∈Xx\in X. The infimum of such bounds is called the essential supremum of ∣f∣|f|, denoted by
∥f∥∞:=ess sup⁡x∈X∣f(x)∣=inf⁡{M≥0:∣f(x)∣≤M for μ-almost every x∈X}.\begin{equation*}\|f\|_{\infty} := \operatorname{ess\,sup}_{x\in X}|f(x)| = \inf\{M\geq 0: |f(x)|\leq M \text{ for }\mu\text{-almost every }x\in X\}.\end{equation*}

Hölder and Minkowski inequalities

This section covers two most important inequality for Lp(X,μ)L_p(X, \mu) spaces: Hölder's and Minkowski's inequalities.

Hölder inequality

Convexity of exp⁡(x)\exp(x)
HÖLDER INEQUALITY
Suppose that 1≤p<∞1 \leq p < \infty and q=pp−1q = \frac{p}{p-1}. If f∈Lp(X,μ)f \in L_p(X, \mu) and g∈Lq(X,μ)g \in L^q(X, \mu), then fg∈L1(X,μ)fg \in L^1(X, \mu). Moreover,
∫X∣f(x)g(x)∣ dμ(x)≤(∫X∣f(y)∣p dμ(y))1/p⋅(∫X∣g(z)∣q dμ(z))1/q.\begin{equation*}\int_X |f(x)g(x)| \, d\mu(x) \leq \left( \int_X |f(y)|^p \, d\mu(y) \right)^{1/p} \cdot \left( \int_X |g(z)|^q \, d\mu(z) \right)^{1/q}.\end{equation*}
Proof. Let a,b>0a, b > 0 and let A=ln⁡(ap)A = \ln(a^p) and B=ln⁡(bq)B = \ln(b^{q}). Since the exponential function is strictly convex,
exp⁡(Ap+Bq)≤exp⁡Ap+exp⁡Bq,\begin{equation*}\exp \left( \frac{A}{p} + \frac{B}{q} \right) \leq \frac{\exp A}{p} + \frac{\exp B}{q} ,\end{equation*}
with equality only if A=BA = B. Hence
exp⁡(ln⁡(ap)p+ln⁡(bq)q)=ab≤app+bqq.\begin{equation*}\exp \left( \frac{\ln(a^p)}{p} + \frac{\ln(b^q)}{q} \right) = ab \leq \frac{a^p}{p} + \frac{b^{q}}{q}.\end{equation*}
The function fgfg is defined a.e. and measurable. If we set a=∣f(x)∣/∥f∥pa = |f(x)| / \|f\|_p and b=∣g(x)∣/∥g∥qb = |g(x)| / \|g\|_q then by previous inequality we get
∣f(x)g(x)∣∥f∥p∥g∥q≤1p∣f(x)∣p∥f∥pp+1q∣g(x)∣q∥g∥qq.\begin{equation*}\frac{|f(x)g(x)|}{\|f\|_p\|g\|_q} \leq \frac{1}{p} \frac{|f(x)|^p}{\|f\|_p^p} + \frac{1}{q} \frac{|g(x)|^q}{\|g\|_q^q}.\end{equation*}
Integrating we get
∫X∣f(x)g(x)∣∥f∥p∥g∥q dμ(x)≤1p∫X∣f(x)∣p∥f∥pp dμ(x)+1q∫X∣g(x)∣q∥g∥qq dμ(x)=1p+1q=1.\begin{equation*}\int_X \frac{|f(x)g(x)|}{\|f\|_p\|g\|_q} \, d\mu(x)\leq\frac{1}{p} \int_X \frac{|f(x)|^p}{\|f\|_p^p} \, d\mu(x)+\frac{1}{q} \int_X \frac{|g(x)|^q}{\|g\|_q^q} \, d\mu(x)= \frac{1}{p} + \frac{1}{q} = 1.\end{equation*}
This is exactly Hölder's inequality. □\Box
An immediate corollary of Hölder’s inequality is the following Cauchy–Schwarz inequality
CAUCHY–SCHWARZ INEQUALITY
If f∈L2(μ)f \in L^2(\mu), g∈L2(μ)g \in L^2(\mu). Then fg∈L1(μ)fg \in L^1(\mu) and
∫Xf(x)g(x) dμ(x)≤(∫X∣f(y)∣2 dμ(y))1/2(∫X∣g(z)∣2 dμ(z))1/2.\begin{equation*}\int_X f(x)g(x) \, d\mu(x) \leq \left( \int_X |f(y)|^2 \, d\mu(y) \right)^{1/2} \left( \int_X |g(z)|^2 \, d\mu(z) \right)^{1/2}.\end{equation*}

Minkowski’s inequality

To establish the Minkowski inequalities, we first introduce the following simple lemma.
Convexity of x4x^4
LEMMA
If 1≤p<∞1 \leq p<\infty and a,b≥0a, b \geq 0, then
(a+b)p≤2p−1(ap+bp).\begin{equation*}(a+b)^p \leq 2^{p-1}\left(a^p+b^p\right) .\end{equation*}
Proof. If p=1p=1, then equality holds. For p>1p>1, the function tpt^p is convex on [0,∞)[0, \infty); that is, its graph lies below the chord line joining the points (a,ap)\left(a, a^p\right) and (b,bp)\left(b, b^p\right). Thus
(a+b2)p≤ap+bp2,\begin{equation*}\left(\frac{a+b}{2}\right)^p \leq \frac{a^p+b^p}{2},\end{equation*}
from which the statement of lemma follows at once. □\Box
MINKOWSKI’S INEQUALITY
Let 1≤p<∞ 1 \leq p < \infty and let f,g∈Lp(X,μ)f, g \in L_p(X, \mu). Then f+g∈Lp(μ)f + g \in L_p(\mu) and
(∫X∣f(x)+g(x)∣p dμ(x))1/p≤(∫X∣f(x)∣p dμ(x))1/p+(∫X∣g(x)∣p dμ(x))1/p\begin{equation*}\left( \int_X |f(x) + g(x)|^p \, \text{d} \mu(x) \right)^{1/p} \leq \left( \int_X |f(x)|^p \, \text{d} \mu(x) \right)^{1/p} + \left( \int_X |g(x)|^p \, \text{d} \mu(x) \right)^{1/p}\end{equation*}
or in the norm form
∥f+g∥Lp(X,μ)≤∥f∥Lp(X,μ)+∥g∥Lp(X,μ).\begin{equation*}\|f + g\|_{L_p(X, \mu)} \leq \|f\|_{L_p(X, \mu)} + \|g\|_{L_p(X, \mu)} .\end{equation*}
\quad Proof.Consider the simple inequality
∣f+g∣p=∣f+g∣⋅∣f+g∣p−1≤∣f∣⋅∣f+g∣p−1+∣g∣⋅∣f+g∣p−1,\begin{equation*}|f + g|^p = |f+g| \cdot |f+g|^{p-1} \leq |f|\cdot|f+g|^{p-1} + |g|\cdot|f+g|^{p-1},\end{equation*}
and integrate over XX, then apply Hölder's inequality to obtain
∫X∣f(x)+g(x)∣p dμ(x)≤∫X∣f(x)∣⋅∣f(x)+g(x)∣p−1 dμ(x)+∫X∣g(x)∣⋅∣f(x)+g(x)∣p−1 dμ(x)≤Ho¨lder’s(∥f∥Lp(X,μ)+∥g∥Lp(X,μ))⋅∥∣f+g∣p−1∥Lq(X,μ).\begin{align*}&\int_X |f(x) + g(x)|^p \, \text{d} \mu(x) \leq \\&\quad \int_X |f(x)| \cdot |f(x) + g(x)|^{p-1} \, \text{d} \mu(x)+ \int_X |g(x)| \cdot |f(x) + g(x)|^{p-1} \, \text{d} \mu(x) \\&\overset{\text{Hölder's}}{\leq} \left(\|f\|_{L_p(X,\mu)} + \|g\|_{L_p(X,\mu)}\right) \cdot \left\||f+g|^{p-1}\right\|_{L^q(X,\mu)}.\end{align*}
Since
q=pp−1⟹∥∣f+g∣p−1∥Lq(X,μ)=(∫X∣f(x)+g(x)∣pdμ(x))1q\begin{equation*}q = \frac{p}{p-1} \quad \Longrightarrow \quad \left\| | f + g |^{p-1}\right\|_{L^q(X, \mu)} = \left( \int_X | f(x) + g(x) |^{p} \text{d} \mu(x) \right)^{\frac{1}{q}}\end{equation*}
from which we obtain by dividing previous inequality by (∫X∣f+g∣p dμ)1/q\left( \int_X |f + g|^p \, d\mu \right)^{1/q}
(∫X∣f(x)+g(x)∣p dμ(x))1−1q≤∥f∥Lp(X,μ)+∥g∥Lp(X,μ)⟹1−1q=1p∥f+g∥p≤∥f∥p+∥g∥p.□\begin{equation*}\left( \int_X |f(x) + g(x)|^p \, \text{d} \mu(x) \right)^{1 - \frac{1}{q}} \leq \| f \|_{L_p(X, \mu)} + \| g \|_{L_p(X, \mu)}\overset{1-\frac{1}{q}=\frac{1}{p}}{\Longrightarrow}\|f + g\|_p \leq \|f\|_p + \|g\|_p. \qquad \Box\end{equation*}
THEOREM
For 1≤p≤∞1 \leq p \leq \infty, Lp(X,μ)L_p(X, \mu) is a normed linear space
Proof. The Minkowski's inequality means that, whenever 1≤p<∞1 \leq p < \infty, the formula
d(f,g):=∥f−g∥p\begin{equation*}d(f, g) := \|f - g\|_p\end{equation*}
defines a metric on Lp(μ)L_p(\mu), since the triangle inequality holds true
∥g−h∥p=∥g−f+f−h∥p≤∥g−f∥p+∥f−h∥p.□\begin{equation*}\|g - h\|_p = \|g - f + f - h\|_p \leq \|g - f\|_p + \|f - h\|_p. \Box\end{equation*}

Completeness of Lp(X,μ)L_p(X,\mu)

This is Theorem 2.16 from Adams and Fournier [AdamsFournier2003].
We want to show that the spaces Lp(μ)L_p(\mu), where 1≤p≤∞1 \leq p \leq \infty are Banach spaces. In order to do it we need to prove that these spaces are complete, e.g. for every Cauchy sequences {fn}\left\{f_n\right\}
∀ε>0,  ∃N such that ∀n,m≥N:∥fn−fm∥p=(∫X∣fm−fn∣p dμ)1p<ε,\begin{equation*}\forall \varepsilon > 0, \; \exists N \text{ such that } \forall n, m \geq N: \|f_n - f_m\|_p = \left( \int_X | f_m - f_n |^p \, d\mu \right)^{\frac{1}{p}} < \varepsilon ,\end{equation*}
exist a limit function inside Lp(μ)L_p(\mu).
THEOREM
The space Lp(X,μ)L_p(X,\mu), where 1≤p≤∞1 \leq p \leq \infty, is complete with respect to the metric ∥f−g∥p\|f - g\|_p.

The Cauchy sequence

First assume 1≤p<∞1 \leq p<\infty and let {fn}\left\{f_n\right\} be a Cauchy sequence in Lp(μ)L_p(\mu). We can extract subsequence {fnj}\left\{f_{n_j}\right\} of {fn}\left\{f_n\right\} such that
∥fnj+1−fnj∥p≤12j,j=1,2,…\begin{equation*}\| f_{n_{j+1}}-f_{n_j}\|_p \leq \frac{1}{2^j}, \quad j=1,2, \ldots\end{equation*}

The convergent monotone sequence

Let's define the monotonicly increasing sequence {gn}n=1∞\{ g_n\}^{\infty}_{n=1} for x∈Xx \in X as
gm(x)=∑j=1m∣fnj+1(x)−fnj(x)∣⟹0≤g1≤g2≤⋯≤gm≤…\begin{equation*}g_m(x) = \sum_{j=1}^m | f_{n_{j+1}}(x)-f_{n_j}(x) | \quad \Longrightarrow \quad 0 \leq g_1 \leq g_2 \leq \dots \leq g_m \leq \dots\end{equation*}
and due to Minkowski's inequality for any mm
∫Xgmp(x)dμ(x)≤(∥f1∥Lp(X,μ)+∑j=1m∥fnj+1−fnj∥Lp(X,μ))p≤1,\begin{equation*}\int_X g^p_m(x) \text{d} \mu(x) \leq \left( \| f_1 \|_{L_p(X, \mu)} + \sum_{j=1}^m \| f_{n_{j+1}}-f_{n_j} \|_{L_p(X, \mu)} \right)^{p} \leq 1 ,\end{equation*}
we get that {gn}\{ g_n\} is bounded in Lp(X,μ)L_p(X, \mu). Putting gp(x)=lim⁡m→∞gmp(x)g^p(x) = \lim_{m \rightarrow \infty} g^p_m(x), we obtain
gmp(x)↑gp(x)⟹Monotone Convergence Theorem∫Xgp(x)dμ(x)=lim⁡m→∞∫Xgmp(x)dμ(x)≤1.\begin{equation*}g_m^p(x)\uparrow g^p(x)\overset{\text{Monotone Convergence Theorem}}{\Longrightarrow}\int_X g^p(x) \text{d} \mu(x) = \lim_{m \rightarrow \infty} \int_X g^p_m(x) \text{d} \mu(x) \leq 1.\end{equation*}
Hence gp(x)<∞g^p(x) < \infty μ−\mu-a.e on XX and g∈Lp(X,μ) g \in L_p(X, \mu).

Pointwise convergence of {fnj}\{ f_{n_j}\}

Our next goal is to prove that {fnj(x)}\{f_{n_j}(x)\} is a Cauchy sequence for almost every x∈Xx\in X, and hence converges pointwise almost everywhere. For all k≥1,k \geq 1,
∣fnm+k−fnm∣=∣fnm+k−fnm+k−1+fnm+k−1−⋯+fnm+1−fnm∣≤∑i=m+1k∣fni−fni−1∣=gm+k−gm→0μ−a.e.\begin{equation*}| f_{n_{m+k}}-f_{n_m} | = | f_{n_{m+k}} - f_{n_{m+k-1}} + f_{n_{m+k-1}} - \dots + f_{n_{m+1}}- f_{n_m} | \leq \sum_{i=m+1}^{k} | f_{n_{i}}-f_{n_{i-1}} | = g_{m+k} - g_m \rightarrow 0 \quad \mu-\text{a.e.}\end{equation*}
Therefore, {fnm(x)}m=1∞ \{ f_{n_{m}}(x) \}_{m = 1}^{\infty} is Cauchy sequence on real line, so it converges. Now we can define the limit
f(x)=lim⁡m→∞fnm(x)μ−a.e. on X\begin{equation*}f(x)=\lim_{m \rightarrow \infty} f_{n_m} (x) \quad \mu-\text{a.e. on } X\end{equation*}
Since ∣fnm∣≤gm≤g|f_{n_m}| \leq g_m \leq g μ−\mu-a.e, ∣f∣≤g|f| \leq g and ∣fnm−f∣≤2pgp|f_{n_m} - f| \leq 2^p g^p. Then by Dominated Convergence Theorem,
lim⁡m→∞∫X∣f−fnm∣pdμ(x)=∫Xlim⁡m→∞∣f−fnm∣pdμ(x)=0\begin{equation*}\lim_{m \rightarrow \infty} \int_{X} | f - f_{n_m} |^p \text{d} \mu(x) = \int_{X} \lim_{m \rightarrow \infty} | f - f_{n_m} |^p \text{d} \mu(x) = 0\end{equation*}

Convergence to ff

For any ϵ>0\epsilon>0 there exists NN such that if m,n≥Nm, n \geq N, then ∥fm−fn∥p<ϵ\left\|f_m-f_n\right\|_p<\epsilon. Hence, by Fatou's lemma
∫X∣f(x)−fn(x)∣pdμ(x)=∫Xlim⁡j→∞∣fnj(x)−fn(x)∣pdμ(x)≤lim inf⁡j→∞∫X∣fnj(x)−fn(x)∣pdμ(x)≤ϵp.\begin{align*}\int_{X}\left|f(x)-f_n(x)\right|^p \text{d} \mu(x) & =\int_{X} \lim _{j \rightarrow \infty}\left|f_{n_j}(x)-f_n(x)\right|^p \text{d} \mu(x) \\& \leq \liminf _{j \rightarrow \infty} \int_{X}\left|f_{n_j}(x)-f_n(x)\right|^p \text{d} \mu(x) \leq \epsilon^p.\end{align*}
Thus ∥f−fn∥p→0\left\|f-f_n\right\|_p \rightarrow 0 as n→∞n \rightarrow \infty. Therefore Lp(Ω)L_p(\Omega) is complete and so is a Banach space.

Completeness of L∞(X,μ)L^{\infty}(X, \mu)

Finally, if {fn}\left\{f_n\right\} is a Cauchy sequence in L∞(X,μ)L^{\infty}(X, \mu), then there exists a set A⊂XA \subset X having measure zero such that if x∉Ax \notin A, then for every n,m=1,2,…n, m=1,2, \ldots
∣fn(x)∣≤∥fn∥∞,∣fn(x)−fm(x)∣≤∥fn−fm∥∞.\begin{equation*}\left|f_n(x)\right| \leq\left\|f_n\right\|_{\infty}, \quad\left|f_n(x)-f_m(x)\right| \leq\left\|f_n-f_m\right\|_{\infty} .\end{equation*}
Therefore, {fn}\left\{f_n\right\} converges uniformly on X/AX / A to a bounded function ff. Setting f=0f=0 for x∈Ax \in A, we have u∈L∞(Ω)u \in L^{\infty}(\Omega) and ∥fn−f∥∞→0\left\|f_n-f\right\|_{\infty} \rightarrow 0 as n→∞n \rightarrow \infty. Thus L∞(X,μ)L^{\infty}(X, \mu) is also complete and a Banach space. □\Box

Approximation Lp(Rn)L_p(\mathbb{R}^n) by Simple and Continuous Functions

This is Theorem 2.19 from Adams and Fournier [AdamsFournier2003].

Simple functions

Approximation is one of the basic tools in LpL_p theory. Many statements are first proved for very simple functions and then extended to general LpL_p functions by density. We begin with simple functions.
THEOREM
If p∈[1,∞) p \in [1, \infty) , then the set of simple functions f=∑i=1nai1Ei f = \sum_{i=1}^n a_i \mathbf{1}_{E_i} , where each Ei E_i is an element of the σ\sigma-algebra X \mathcal{X} and μ(Ei)<∞ \mu(E_i) < \infty , is dense in Lp(X,X,μ) L_p(X, \mathcal{X}, \mu) .
\quad Proof. If f∈Lp f \in L_p , then f f is measurable; thus, there exists a sequence {ϕn}n=1∞ \{\phi_n\}_{n=1}^\infty of simple functions, such that ϕn→f \phi_n \rightarrow f almost everywhere with
0≤∣ϕ1∣≤∣ϕ2∣≤⋯≤∣f∣,\begin{equation*}0 \leq |\phi_1| \leq |\phi_2| \leq \cdots \leq |f|,\end{equation*}
i.e., ϕn \phi_n approximates f f from below. For approximation sequence we get ∣ϕn−f∣p→0 |\phi_n - f|_p \rightarrow 0 μ\mu-a.e. and ∣ϕn−f∣≤2∣f∣ |\phi_n - f| \leq 2|f| Since ∣f∣p∈L1(μ)|f|^p \in L^1(\mu), the Dominated Convergence Theorem gives
∥ϕn−f∥pp=∫X∣ϕn−f∣p dμ→0.⟹∥ϕn−f∥p→0.□\begin{equation*}\|\phi_n-f\|_p^p = \int_X |\phi_n-f|^p\,d\mu \to 0. \quad \Longrightarrow \quad \|\phi_n-f\|_p \to 0. \quad \Box\end{equation*}

Continuous functions C0(Rn)C_0(\mathbb{R}^n)

Simple functions are useful for measure-theoretic arguments, but in analysis it is often more convenient to work with continuous functions. On Rn\mathbb R^n, every LpL_p function can in fact be approximated by continuous functions with compact support.
THEOREM
C0(Rn)C_0(\mathbb{R}^n) is dense in Lp(Rn)L_p(\mathbb{R}^n) if 1≤p<∞1 \leq p<\infty.
\quad Proof. Let u∈Lp(Rn)u \in L_p(\mathbb{R}^n). As we have seen in the previous theorem, there exists a nonnegative sequence of simple functions {ϕn}\{\phi_n\} and some ϕ∈{ϕn}\phi \in \{\phi_n\} such that ∥u−ϕ∥p<ϵ/2\|u-\phi\|_p<\epsilon / 2. Since ϕ\phi has compact support, it vanishes outside some bounded measurable set EE of finite measure. By Lusin's theorem, there exists γ∈Cc(Rn)\gamma \in C_c(\mathbb R^n) such that
∣γ(x)∣≤∥ϕ∥∞for all x,\begin{equation*}|\gamma(x)|\le \|\phi\|_\infty \quad \text{for all } x,\end{equation*}
and
vol⁡{x:γ(x)≠ϕ(x)}<(ε4∥ϕ∥∞)p.\begin{equation*}\operatorname{vol}\{x:\gamma(x)\neq \phi(x)\}<\left(\frac{\varepsilon}{4\|\phi\|_\infty}\right)^p.\end{equation*}
Therefore
∥ϕ−γ∥p≤∥ϕ−γ∥∞ vol⁡{x:γ(x)≠ϕ(x)}1/p<Lusin’s theorem2∥ϕ∥∞⋅ε4∥ϕ∥∞=ε2.\begin{equation*}\|\phi-\gamma\|_p\le\|\phi-\gamma\|_\infty\,\operatorname{vol}\{x:\gamma(x)\neq \phi(x)\}^{1/p}\overset{\text{Lusin's theorem}}{<}2\|\phi\|_\infty \cdot \frac{\varepsilon}{4\|\phi\|_\infty}=\frac{\varepsilon}{2}.\end{equation*}
Hence
∥u−γ∥p≤∥u−ϕ∥p+∥ϕ−γ∥p<ε.□\begin{equation*}\|u-\gamma\|_p \le \|u-\phi\|_p + \|\phi-\gamma\|_p < \varepsilon. \quad \Box\end{equation*}

Separability of Lp(Rn)L_p(\mathbb{R}^n)

This is Theorem 2.21 from Adams and Fournier [AdamsFournier2003].
In this section we assume that Ω⊂Rn\Omega \subset \mathbb R^n is open and that μ\mu is Lebesgue measure.
THEOREM
If Ω⊂Rn\Omega \subset \mathbb R^n is open and 1≤p<∞1 \le p < \infty, then Lp(Ω)L_p(\Omega) is separable.
\quad Proof. For m=1,2,…m=1,2, \ldots let
Ωm={x∈Ω:∣x∣≤m and dist⁡(x,bdry⁡(Ω))≥1/m}.\begin{equation*}\Omega_m=\{x \in \Omega:|x| \leq m \text { and } \operatorname{dist}(x, \operatorname{bdry}(\Omega)) \geq 1 / m\} .\end{equation*}
Then Ωm\Omega_m is a compact subset of Ω\Omega. Let PP be the set of all polynomials on Rn\mathbb{R}^n having rational-complex coefficients, and let Pm={1Ωmf:f∈P}P_m=\left\{\mathbf{1}_{\Omega_m} f: f \in P\right\}. PmP_m is dense in C(Ωm)C\left(\Omega_m\right). Moreover, ⋃m=1∞Pm\bigcup_{m=1}^{\infty} P_m is countable. If u∈Lp(Ω)u \in L_p(\Omega) and ϵ>0\epsilon>0, there exists ϕ∈C0(Ω)\phi \in C_0(\Omega) such that ∥u−ϕ∥p<ϵ/2\|u-\phi\|_p<\epsilon / 2. If 1/m<dist⁡(supp⁡(ϕ),bdry⁡(Ω))1 / m<\operatorname{dist}(\operatorname{supp}(\phi), \operatorname{bdry}(\Omega)), then there exists ff in the set PmP_m such that ∥ϕ−f∥∞<(ϵ/2)(vol⁡(Ωm))−1/p\|\phi-f\|_{\infty}<(\epsilon / 2)\left(\operatorname{vol}\left(\Omega_m\right)\right)^{-1 / p}. It follows that
∥ϕ−f∥p≤∥ϕ−f∥∞(vol⁡(Ωm))1/p<ϵ/2\begin{equation*}\|\phi-f\|_p \leq\|\phi-f\|_{\infty}\left(\operatorname{vol}\left(\Omega_m\right)\right)^{1 / p}<\epsilon / 2\end{equation*}
and so ∥u−f∥p<ϵ\|u-f\|_p<\epsilon. Thus the countable set ⋃m=1∞Pm\bigcup_{m=1}^{\infty} P_m is dense in Lp(Ω)L_p(\Omega) and Lp(Ω)L_p(\Omega) is separable. □\Box

References

  • [AdamsFournier2003]Adams, Robert A., and John J. F. Fournier. Sobolev Spaces. Second edition. Academic Press, 2003.